Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{-10q - 40}{6q + 42} \times \dfrac{q^2 + 8q + 7}{q + 1} $
Solution: First factor the quadratic. $r = \dfrac{-10q - 40}{6q + 42} \times \dfrac{(q + 7)(q + 1)}{q + 1} $ Then factor out any other terms. $r = \dfrac{-10(q + 4)}{6(q + 7)} \times \dfrac{(q + 7)(q + 1)}{q + 1} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -10(q + 4) \times (q + 7)(q + 1) } { 6(q + 7) \times (q + 1) } $ $r = \dfrac{ -10(q + 4)(q + 7)(q + 1)}{ 6(q + 7)(q + 1)} $ Notice that $(q + 1)$ and $(q + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -10(q + 4)\cancel{(q + 7)}(q + 1)}{ 6\cancel{(q + 7)}(q + 1)} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $r = \dfrac{ -10(q + 4)\cancel{(q + 7)}\cancel{(q + 1)}}{ 6\cancel{(q + 7)}\cancel{(q + 1)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $r = \dfrac{-10(q + 4)}{6} $ $r = \dfrac{-5(q + 4)}{3} ; \space q \neq -7 ; \space q \neq -1 $